Chapter 10, Problem 12CR

Find the rectangle equation of the plane curve

$x(t)=5\tan ty(t)=5{\sec }^{2}t-\frac{\pi }{2}<t<\frac{\pi }{2}$

To determine

The rectangular equation of the plane curve, if $x\left(t\right)=5\tan t,y\left(t\right)=5{\sec }^{2}t$ and $-\frac{\pi }{2}<t<\frac{\pi }{2}$

Answer to Problem 12CR

Solution:

The rectangular equation of the plane curve $x\left(t\right)=5\tan t,y\left(t\right)=5{\sec }^{2}t$ is $x^2=5\left(y-5\right)$.

Explanation of Solution

Given information:

$x\left(t\right)=5\tan t,y\left(t\right)=5{\sec }^{2}t$ and $-\frac{\pi }{2}<t<\frac{\pi }{2}$

Explanation:

Now, let $x=5\tan t$ and $y=5{\sec }^{2}t$

$\Rightarrow \frac{x}{5}=\tan t$ and $\frac{y}{5}={\sec }^{2}t$

$\Rightarrow {\left(\frac{x}{5}\right)}^2={\tan }^{2}t$

$\Rightarrow \frac{x^2}{25}={\tan }^{2}t$

By using identity ${\tan }^{2}t+1={\sec }^{2}t$,

$\Rightarrow \frac{x^2}{25}+1=\frac{y}{5}$

$\Rightarrow x^2+25=\frac{y}{5}\cdot 25$

$\Rightarrow x^2+25=5y$

$\Rightarrow x^2=5\left(y-5\right)$

The equation $x^2=5\left(y-5\right)$ represents a parabola.

Therefore, the rectangular equation of the plane curve, $x\left(t\right)=5\tan t,y\left(t\right)=5{\sec }^{2}t$ is $x^2=5\left(y-5\right)$